Question

if root3+i whole power 100=2 power99(a+ib) then show that a square +b square =4​

Answer 1

Answer:

Step-by-step explanation: Given

$( \sqrt{3} +i)^{100} =2^{99} (a+ib)$

$(\frac{ \sqrt{3} +i}{2} )^{100} =2^{-1} (a+ib)$

$w^{100} =2^{-1} (a+ib)\\(w^{3} )^{33} w=2^{-1} (a+ib)\\w=2^{-1} (a+ib)\\\frac{\sqrt{3} +i}{2} =2^{-1} (a+ib)$

$\frac{\sqrt{3} -i}{2} =2^{-1} (a-ib)$

$(\frac{\sqrt{3} -i}{2}) (\frac{\sqrt{3}+i}{2})=2^{-1} (a-ib)2^{-1} (a+ib)$

$\frac{3+1}{4} =2^{-2}(a^{2} +b^{2} )\\a^{2} +b^{2} =4$